Problems

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Solution: \begin{align*} && 1 &= x^2 + y^2 + 2axy \\ \frac{\d}{\d x}: && 0 &= 2x + 2y \frac{\d y}{\d x} + 2ay + 2ax \frac{\d y}{\d x} \\ &&&= (2x+2ay) + \frac{\d y}{\d x} \left (2ax + 2y \right) \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{x+ay}{ax+y} \end{align*}

The gradient of \(OP\) is \(\frac{y}{x}\). The gradient of the normal is \(\frac{ax+y}{x+ay}\) Therefore (noting the absolute values in case they are on opposite sides to this diagram: \begin{align*} && \tan \theta &= \Big |\tan \left ( \tan^{-1} \frac{ax+y}{x+ay} - \tan^{-1} \frac{y}{x} \right) \Big | \\ &&&= \Big | \frac{\frac{ax+y}{x+ay} - \frac{y}{x}}{1+\frac{ax+y}{x+ay}\frac{y}{x} } \Big | \\ &&&= \Big | \frac{(ax+y)x - y(x+ay)}{x(x+ay)+y(ax+y)} \Big | \\ &&&= \Big | \frac{ax^2 - ay^2}{x^2+y^2+2ayx} \Big | \\ &&&= a \frac{|y^2-x^2|}{1} \\ &&&= a|y^2-x^2| \end{align*}

1. \(\,\) \begin{align*} && \sec^2 \theta \frac{\d \theta}{\d x} &= \pm a \left (2y \frac{\d y}{\d x} - 2 x\right) \\ \Rightarrow && 0 &= a \left (y \cdot \frac{x+ay}{ax+y} + x \right) \\ &&&=a \left ( \frac{xy+ay^2+ax^2+xy}{ax+y} \right) \\ \Rightarrow && 0 &= a(x^2+y^2)+2xy \end{align*}
2. \(\,\) \begin{align*} && 0 &=a(x^2+y^2)+2xy \\ && 1 &= x^2+y^2 + 2axy \\ \Rightarrow && 1 &= (a+1)(x^2+y^2) + (a+1)(2xy) \\ &&&= (a+1)(x^2+y^2+2xy) \end{align*}
3. \(\,\) \begin{align*} && 1 &= (a+1)(x+y)^2 \\ \Rightarrow && x +y &= \pm \frac{1}{\sqrt{1+a}} \\ && 0 &=a(x^2+y^2)+2xy \\ && 1 &= x^2+y^2 + 2axy \\ \Rightarrow && 1 &= (1-a)(x^2+y^2) + (a-1)(2xy) \\ &&&= (1-a)(x^2+y^2-2xy)\\ \Rightarrow && x-y &= \pm \frac{1}{\sqrt{1-a}} \\ \Rightarrow && \frac{\d \theta}{\d x} &= a|y^2-x^2| \\ &&&= a|(y-x)(x+y)| \\ &&&= \frac{a}{\sqrt{1-a^2}} \end{align*}

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Solution: \begin{align*} && \frac{\d y}{\d x} &= - \frac{xy}{x^2+a^2} \\ \Rightarrow && \int \frac{1}{y} \d y &= \int -\frac{x}{x^2+a^2} \d x \\ && \ln |y| &= -\frac12 \ln |x^2 + a^2| + C \\ \Rightarrow && C' &= \ln y^2 + \ln (x^2+a^2) \\ \Rightarrow && c^2 &= y^2(x^2+a^2) \end{align*} (where the final constant \(c^2\) can be taken as a square since it is clearly positive).

\begin{align*} && \frac{\d }{\d x} \left (x^2 + y^2 \right) &= 2x - \frac{2xy^2}{x^2+a^2} \\ &&&=2x - \frac{2x c^2}{(x^2+a^2)^2} \\ &&&= 2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \\ \\ && \frac{\d ^2}{\d x^2} \left (x^2 + y^2 \right) &= \frac{\d }{\d x} \left (2x \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + 2x \left (\frac{2c^2 \cdot 2x}{(x^2+a^2)^3} \right) \\ &&&= 2 \left ( 1 - \frac{c^2}{(x^2+a^2)^2}\right) + \frac{8x^2c^2 }{(x^2+a^2)^3} \\ \end{align*}

1. The shortest distance from the origin will have the first derivative as \(0\), ie \(x = 0\) or \(x^2 + a^2 = c\), but if \(c < a^2\) this can only occur for \(x = 0\), so the closest to the origin is \((0, \pm \frac{c}{a})\)
2. If \(c > a^2\) then we can have \(x = 0\) or \(x = \pm \sqrt{c-a^2}\). Looking at the second derivative, when \(x = 0\) we have \(2(1- \frac{c^2}{a^4}) < 0\) which is a local maximum. When \(x = \pm\sqrt{c-a^2}\) we have \(8(c-a^2)c^2/c^3 > 0\) which is the minimum, therefore the points are \((\pm \sqrt{c-a^2}, c)\)

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Solution:

1. \begin{align*} && c &= \left (y+2x \right)^3\left (y-4x \right) \\ \Rightarrow && 0 &= 3\left (y+2x \right)^2\left (y-4x \right)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right)^3 \left ( \frac{\d y}{\d x} - 4 \right) \\ \Rightarrow && 0 &= 3(y-4x)\left ( \frac{\d y}{\d x} + 2 \right) + \left (y+2x \right) \left ( \frac{\d y}{\d x} - 4 \right) \\ \Rightarrow &&&= \frac{\d y}{\d x} \left (3(y-4x) + (y+2x) \right) + 6(y-4x)-4(y+2x) \\ &&&= \frac{\d y}{\d x} \left ( 4y-10x\right) + 2y-32x \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{16x-y}{2y-5x} \end{align*}
2. \begin{align*} && c &= \left ( y + ax \right)^n \left ( y + bx \right) \\ \Rightarrow && 0 &= n\left ( y + ax \right)^{n-1} \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right)^{n} \left ( \frac{\d y}{ \d x}+b \right) \\ \Rightarrow && 0 &= n \left ( y + bx \right)\left ( \frac{\d y}{ \d x}+a \right) + \left ( y + ax \right) \left ( \frac{\d y}{ \d x}+b \right) \\ &&&= \frac{\d y}{\d x} \left ( (n+1)y + (nb+a)x \right) + an(y+bx) + by+bax \\ &&&= \frac{\d y}{\d x} \left ( (n+1)y + (nb+a)x \right) + (an+b)y+ab(n+1)x \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{(an+b)y+ab(n+1)x}{(n+1)y+(nb+a)x} \end{align*} We must have \(ab = 10, a+b = -7\) so say \(a=-5,b=-2,n=2\) and we have \((y-5x)^2(y-2) = c\) is our general solution to the differential equation

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Solution: \begin{align*} && y & = \ln ( x + \sqrt{x^2+1}) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{x+\sqrt{x^2+1}} \cdot \frac{\d }{\d x} \left ( x + \sqrt{x^2+1} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left (1 + \frac12 \frac{2x}{\sqrt{x^2+1}} \right) \\ &&&= \frac{1}{x+\sqrt{x^2+1}} \left ( \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}\right) \\ &&&= \frac{1}{\sqrt{x^2+1}} \end{align*} Note that \(\displaystyle y^{(2)} = - \frac12 \frac{2x}{(x^2+1)^{3/2}} = - \frac{x}{(x^2+1)^{3/2}}\), and in particular \((x^2+1)y^{(2)} + xy^{(1)} = 0\). Now applying Leibnitz formula: \begin{align*} 0 &= \left ( (x^2+1)y^{(2)} + xy^{(1)} \right )^{(n)} \\ &= \left ( (x^2+1)y^{(2)}\right )^{(n)} + \left (xy^{(1)} \right )^{(n)} \\ &= (x^2+1)y^{(n+2)} +n2xy^{(n+1)} + \binom{n}{2}2y^{(n)} + xy^{(n+1)} + n y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + (n^2-n+n)y^{(n)} \\ &= (x^2+1)y^{(n+2)} + (2n+1)xy^{(n+1)} + n^2y^{(n)} \end{align*} as required. When \(x = 0\): \begin{align*} && y(0) &= \ln(0 + \sqrt{0^2+1}) \\ &&&= \ln 1 = 0 \\ && y'(0) &= \frac{1}{\sqrt{0^2+1}} = 1 \\ && y^{(n+2)} &= -n^2 y^{(n)} \\ && y^{(2k)} &= 0 \\ && y^{(3)} &= -1 \\ && y^{(5)} &= 3^2 \\ && y^{(7)} &= - 5^2 \cdot 3^2 \\ \end{align*} Therefore the Maclaurin series about \(x = 0\) is \begin{align*} y &= x - \frac{1}{3!} x^3 + \frac{3^2}{5!} x^5 - \frac{3^2 \cdot 5^2}{7!} x^7 + \cdots \\ &= x - \frac{1}{6} x^3 + \frac{3}{1 \cdot 2 \cdot 4 \cdot 5} x^5 - \frac{5}{1 \cdot 2 \cdot 4 \cdot 2 \cdot 7} x^7 + \cdots \\ &= x - \frac{1}{6}x^3 + \frac{3}{40} x^5 - \frac{5}{56} x^7 + \cdots \end{align*}

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Solution:

1. \begin{align*} && f(x)&=\tan^{-1}x+\tan^{-1}\left(\frac{1-x}{1+x}\right) \\ \Rightarrow && f'(x) &= \frac{1}{1+x^2} + \frac{1}{1+\l \frac{1-x}{1+x} \r^2} \cdot \l \frac{-2}{(1+x)^2}\r \\ &&&= \frac1{1+x^2}- \frac{2}{(1+x)^2+(1-x)^2} \\ &&&= \frac1{1+x^2} - \frac{2}{2+2x^2} \\ &&&= 0 \end{align*} Therefore $f(x) = \begin{cases} c_1 & \text{if } x < -1 \\ c_2 & \text{if } x > -1 \end{cases}$ \(f(0) = \tan^{-1} 0 + \tan^{-1} 1 = \frac{\pi}{4}\) \(\lim_{x \to \infty} f(x) = -\frac{\pi}{2} + \tan^{-1} -1 = -\frac{3\pi}{4}\) therefore $f(x) = \begin{cases} -\frac{3\pi}{4}& \text{if } x < -1 \\ \frac{\pi}{4} & \text{if } x > -1 \end{cases}$
2. \begin{align*} && x &= y \sin y^2 \\ \Rightarrow && \frac{\d x}{\d y} &= \sin y^2 + 2y^2 \cos y^2 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{1}{\sin y^2+2y^2 \cos y^2} \\ &&&=\frac{1}{\frac{x}{y}+2y^2 \sqrt{1-\sin^2y^2}} \\ &&&= \frac{y}{x + 2y^3 \sqrt{1-\frac{x^2}{y^2}}} \\ &&&= \frac{y}{x+2y^2 \sqrt{y^2-x^2}} \end{align*}

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Solution: \begin{align*} && y^2 &= x^2a^2-x^2 \\ &&&= \frac{a^4}{4} -\left ( x^2 -\frac{a^2}{2} \right)^2 \end{align*} Therefore the maximum and minimum values of \(y\) are \(\pm \frac{a^2}2\)

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Solution: \begin{align*} && 0 &= 3y^{2}-10xy+3x^{2}+16y-16x+15 \\ \Rightarrow && 0 &= 6y \frac{\d y}{\d x} - 10x \frac{\d y}{\d x} - 10y + 6x+ 16 \frac{\d y}{\d x } - 16 \\ &&&= \frac{\d y}{\d x} \left (6y - 10x +16 \right) - (10y-6x+16) \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{5y-3x+8}{3y-5x+8} \\ \Rightarrow && \frac{y-t}{x-s} &= \frac{5t-3s+8}{3t-5s+8} \\ && y &= \left(\frac{5t-3s+8}{3t-5s+8}\right)x -\left(\frac{5t-3s+8}{3t-5s+8}\right)s + t \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{(5ts-3s^2+8s)-(3t^2-5st+8t)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(3s^2+3t^2-10st)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8s-8t-(16s-16t-15)}{3t-5s+8} \\ &&&= \left(\frac{5t-3s+8}{3t-5s+8}\right)x - \frac{8t-8s+15}{3t-5s+8} \\ \end{align*} While \(x \to \infty\) we still have \(3 \frac{y^2}{x^2} - 10 \frac{y}{x} + 3 + 16 \frac{y}{x^2} - 16\frac{1}{x} + 15 \frac{1}{x^2} = 0\), ie if \(\frac{y}{x} = k\), then \(3k^2 - 10k + 3 \to 0 \Rightarrow k \to 3, \frac13\). Therefore, as \(s \to \infty\) we can write \begin{align*} && y &= \left(\frac{5\frac{t}{s}-3+8\frac{1}{s}}{3\frac{t}{s}-5+8\frac1{s}}\right)x - \frac{8\frac{t}s-8+15\frac{1}{s}}{3\frac{t}{s}-5+8\frac{1}{s}} \\ k \to 3: &&& \to \left(\frac{15-3}{9-5}\right)x - \frac{24-8}{9-5} \\ &&&= 3x - 4 \\ k \to \frac13: && &\to \left(\frac{\frac53-3}{1-5}\right)x - \frac{\frac83-8}{1-5} \\ &&&= \frac13 x - \frac43 \end{align*} Therefore the equations are \(y = 3x-4\) and \(3y=x-4\) The lines are parallel to \(y = 3x\) and \(y = \frac13x\), so by considering the triangles formed with the origin and a point \(1\) along the \(x\) or \(y\) axis we can see the angles are identical. This means the line \(y = x\) is parallel to one axis and \(y = -x\) is parallel to the other. They must meet where our two lines meet which is \((1,-1)\), so our lines are \(y = x-2\) and \(y = -x\)

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Solution: Note that \(z = xy/(x+y+1) \Rightarrow y(x-z) = z(x+1)\) \begin{align*} && g(x) + g(y) &= g(z) \\ \Rightarrow && g'(x) &= g'(z) \cdot \frac{y(x+y+1) - xy \cdot 1} {(x+y+1)^2} \\ &&&= g'(z) \frac{y^2+y}{(x+y+1)^2} \\ &&&= g'(z) \frac{z^2(y^2+y)}{x^2y^2} \\ &&&= g'(z) \frac{z^2(y+1)}{x^2y} \\ &&&= g'(z) \frac{z^2}{x^2} \left (1 + \frac{x-z}{z(x+1)} \right) \\ &&&= g'(z) \frac{z}{x^2} \frac{zx+x}{x+1} \\ &&&= g'(z) \frac{z(z+1)}{x(x+1)} \end{align*} If \(x = 1\) then as \(y\) ranges from \(0\) to \(\infty\), \(z\) ranges from \(0\) to \(1\), so \(g'(1) = \frac{z(z+1)}{1(1+1)}g'(z)\), ie \(2g'(1) = (u^2+u)g'(u)\). \begin{align*} && g'(u) &= \frac{A}{u(u+1)} \\ \Rightarrow && g(u) &= A\int \left ( \frac{1}{u} - \frac{1}{u+1} \right) \d u \\ &&&= A \left ( \ln u - \ln(u+1) \right) + B \\ &&&= A \ln \left ( \frac{u}{u+1} \right) + B \\ \\ && A \ln \left ( \frac{x}{x+1} \right) + B+A \ln \left ( \frac{y}{y+1} \right) + B &=A \ln \left ( \frac{z}{z+1} \right) + B \\ \Rightarrow && B &= A \ln \left ( \frac{z}{z+1} \frac{y+1}{y} \frac{x+1}{x} \right) \\ &&&= A \ln \left ( \frac{1}{1+\frac{x+y+1}{xy}} \frac{(y+1)(x+1)}{xy} \right) \\ &&&= A \ln 1 \\ &&& = 0 \end{align*} Therefore \(B = 0\). \(A\) cannot be determined from \((*)\). Suppose \(f(x) + f(y) = f(z)\), then \(f'(x) = yf'(z)\). Letting \(x = 1\) we find \(f'(1) = uf'(u) \Rightarrow f(u) = C \ln u + D\), but \(D = 0\) so \(f(x) = C \ln x\)

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Solution: \begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ \frac{\d }{\d x} : && 0 &= y^2+2xyy' + 3x^2+a^2y' \\ \Rightarrow && y' &= -\frac{y^2+3x^2}{a^2+2xy} \\ \\ \frac{\d^2 }{\d x^2}: && 0 &= 2yy'+2yy'+2x(y')^2+2xyy''+6x+a^2y'' \\ \Rightarrow && y'' &= -\frac{4yy'+2x(y')^2+6x}{a^2+2xy} \\ \\ P: && y &= a \\ && y' &= -\frac{a^2}{a^2} = -1 \\ && y'' &= -\frac{-4a}{a^2} = \frac{4}{a} \\ \Rightarrow && y &\approx a - h+\frac{2}{a}h^2 \\ \\ Q: && y &= 0 \\ && y' &= -\frac{3a^2}{a^2} = -3 \\ && y'' &= -\frac{18a+6a}{a^2} = -\frac{24}{a} \\ \Rightarrow && y &\approx 0-3h-\frac{12}{a}h \\ \\ R: && y &= -a \\ && y' &= -\frac{a^2+3a^2}{a^2-2a^2} = 4 \\ && y'' &= -\frac{-16a+32a+6a}{a^2-2a^2} = \frac{22}{a} \\ \Rightarrow && y &\approx -a+4h + \frac{11}{a}h^2 \end{align*} Alternatively: \begin{align*} && 0 &= xy^{2}+x^{3}+a^{2}y-a^{3} \\ P(0,a): && y &\approx a + c_1h + c_2h^2 \\ && 0 &= h(a+c_1h)^2 + a^2(a + c_1h + c_2h^2)-a^3 \\ &&&= a^3-a^3 + (a^2+a^2c_1)h+(2ac_1+a^2c_2)h^2 \\ \Rightarrow && c_1 &=-1, c_2 =\frac{2}{a} \\ \Rightarrow && y &\approx a - h + \frac{2}{a}h^2 \\ \\ Q(a,0): && y &\approx c_1h + c_2h^2 \\ && 0 &= (a+h)(c_1h)^2+(a+h)^3+a^2(c_1h + c_2h^2 )-a^3 \\ &&&= a^3-a^3+(3a^2+a^2c_1)h + (ac_1^2+3a+a^2c_2)h^2 + \cdots \\ \Rightarrow && c_1 &=-3, c_2 = -\frac{12}{a} \\ \Rightarrow && y &\approx -3h -\frac{12}{a}h \\ \\ R(a,-a): && y &\approx -a + c_1h + c_2h^2 \\ && 0 &= (a+h)(-a + c_1h+c_2h^2)^2+(a+h)^3+a^2(-a + c_1h + c_2h^2)-a^3 \\ &&&= (a^2-2a^2c_1+3a^2+a^2c_1)h+(-2ac_1+c_1^2+\cdots)h^2 \\ \Rightarrow && c_1 &=4, c_2 = \frac{11}{a} \\ \Rightarrow && y &\approx -a + 4h + \frac{11}{a} \end{align*}

If \((x,y)\) lies on the curve, then viewing it as a quadratic in \(y\) we must have \(\Delta = (a^2)^2-4\cdot x \cdot (x^3-a^3) \geq 0 \Rightarrow a^4-4x^4+4xa^3 \geq 0 \Rightarrow 4x^4-4a^3x-a^4 \leq 0\)

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Solution:

1. \begin{align*} v &= \mathrm{sech} u \\ &= \frac{1}{\mathrm{cosh } u} \\ &= \frac{1}{\sqrt{1+\mathrm{sinh}^2 u}} \tag{\(u > 0\)} \\ &= \frac{1}{\sqrt{1+\tan^2 \theta}} \\ &= \frac{1}{\sqrt{\mathrm{sec}^2 \theta}} \\ &= \cos \theta \tag{\(0 < \theta < \tfrac{\pi}{2}\)} \end{align*}
2. \begin{align*} && \tan \theta &= \textrm{sinh} u \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && \sec^2 \theta \cdot \frac{\d \theta}{\d u} &= \cosh u \\ \Rightarrow && \frac{\d \theta}{\d u} &=\cosh u \cdot \cos^2 \theta \\ &&&= \frac{1}{v} \cdot v^2 \\ &&&=v \end{align*}
3. \begin{align*} \sin 2 \theta &= 2 \sin \theta \cos \theta \\ &= 2 \sin \theta \cdot \frac{\d \theta}{\d u} \\ &= -2 \frac{\d v}{\d \theta} \cdot \frac{\d \theta}{\d u} \tag{\(\cos \theta = v\)} \\ &= -2 \frac{\d v}{\d u} \end{align*} \begin{align*} && \sin 2 \theta &= -2 \frac{\d v}{\d u} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d u}} && 2 \cos 2 \theta \cdot \frac{\d \theta}{\d u} &= -2 \frac{\d^2 v}{\d u^2} \\ \Rightarrow && \cos 2 \theta &= - \frac{\d^2 v}{\d u^2} \frac{1}{v} \\ &&&= -\frac{\d ^2v}{\d u^2} \cosh u \end{align*}
4. \begin{align*} && \frac{\d u}{\d \theta} &= \frac{1}{v} \\ \Rightarrow && \frac{\d^2 u}{\d \theta^2} &= -\frac{1}{v^2} \frac{\d v}{\d \theta} \\ &&&= \frac{1}{v^2} \sin \theta \\ && \frac{\d v}{\d \theta} &= -\sin \theta \\ \Rightarrow && \frac{\d^2 v}{\d \theta^2} &= -\cos \theta \\ &&&= - v \\ \end{align*} Therefore \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}v}{\mathrm{d}\theta^{2}}+\frac{\mathrm{d}v}{\mathrm{d}\theta}\frac{\mathrm{d}^{2}u}{\mathrm{d}\theta^{2}}+\left(\frac{\mathrm{d}u}{\mathrm{d}\theta}\right)^{2} &= \frac{1}{v} \cdot \left (-v\right) + \left ( - \sin \theta \right ) \cdot \left (\frac{1}{v^2} \sin \theta \right) + \frac{1}{v^2} \\ &= -1 + \frac{1-\sin^2 \theta}{v^2} \\ &= -1 + \frac{\cos^2 \theta}{v^2} \\ &= -1 + 1 \\ &= 0 \end{align*}